- #1
Nelson2436
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Homework Statement
Consider a cannon on a 50m high cliff. The cannon shoots a cannonball with an initial speed of 200m/s at an angle of 30°, causing the cannonball to land 41m away from the cannon. What is the magnitude of the velocity of the cannonball the instance just before it hits the ground?
a) -205m/s
b) 205m/s
c) 173m/s
d) 267m/s
Homework Equations
ax=0 ay=-g
1) Xf=Xi+Vixt
2) Yf=Yi+Viyt-½gt2
3) Vfy=Viy-gt
4) Vfy2=Viy2-2g(Yf-Yi)
The Attempt at a Solution
Vi=200m/s
Vix=200cos30 Viy=200sin30
Xi=0m Yi= 50m
Xf=41m Yf= 0m
In order to find the magnitude of Vf, I need to find its components, Vfx and Vfy. In projectile motion, Vfx=Vix=200cos30, so I just need to find Vfy.
Using equation 1 to find time: t=41/(200cos30)= 0.2367s
I then plugged time into equation 3 to find Vfy=200sin30-(9.8)(0.2367)= 97.68m/s
So Vfx=200cos30= 173.2m/s and Vfy= 97.68m/s so
Vf= √(Vfx2+Vfy2)= √(173.22+97.682)= 199 m/s
I don't see where I'm going wrong, but 199m/s is not one of the 4 multiple choice answers. I would greatly appreciate any help.